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sveticcg [70]
1 year ago
8

Determine the half-life of a nuclide that loses 38.0% of its mass in 407 hours. 590 hours 281 hours 291 hours 568 hour 204 hours

Physics
1 answer:
alexgriva [62]1 year ago
3 0

The half-life of a nuclide is 737 hrs

Let the initial mass of nuclide is N° = 100

The final mass of nuclide after 407 hours

= 100 - 38 = 62

The expression for decay constant is

λ = 1/t ln(N°/Nt)

Substitute the given values and calculate the decay constant as,

λ = 1/407hr ln(100/62)

λ = 9.4 x 10^-4/hr

Now, the half-life is calculated as,

t1/2 = 0.693/λ

= 0.693/9.4 x 10^-4/hr

t1/2 = 737 hrs

Hence, the half-life of a nuclide is 737 hrs

Learn more about Half-Life here:

brainly.com/question/25750315

#SPJ4

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Use the information below to answer questions
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Answer:

The charges are q₁  = 2 × 10⁻⁸ C and  q₂ = 3 × 10⁻⁸ C

Explanation:

Here is the complete question

Two identical tiny balls have charge q1 and q2. The repulsive force one exerts on the other when they are 20cm apart is 1.35 X 10-4 N. after the balls are touched together and then represented once again to 20cm, now the repulsive force is found to be 1.40 X 10-4 N. find the charges q1 and q2.

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The force F = 1.35 × 10⁻⁴ N when the charges are separated a distance of r = 20 cm = 0.2 m is given by

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q₁q₂ = 6 × 10⁻¹⁶ C² (1)

When the charges are brought together, the charge is now q = (q₁ + q₂)/2

The new repulsive force F = 1.406 × 10⁻⁴ N  at a distance of r₂ = 20 cm = 0.2 m is then

F₂ = kq²/r₂²

q² = F₂r₂²/k = 1.406 × 10⁻⁴ N × (0.2 m)²/9 × 10⁹ Nm²/C² = 0.00625 × 10⁻¹³ C² = 6.25 × 10⁻¹⁶ C²

q² = 6.25 × 10⁻¹⁶ C²

q = √(6.25 × 10⁻¹⁶) C

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Expanding the bracket, we have

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So the charges are q₁  = 2 × 10⁻⁸ C and  q₂ = 3 × 10⁻⁸ C

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